Question

. Consider the following model for pastry folding: A piece of pastry is folded in thirds back over itself and then stretched to the original length as illustrated in this diagram This folding and stretching can be modeled by the function f(z) = 2-3r if 1/3 <2/3 with the graph shown below. The diagonal line with 1 # y is shown in red. f(x) (a) (3 marks) Just by looking at the graph, state the number and the positions of the fixed points, that is, the points that do not move when the pastry is folded and stretched (b) (2 marks) Confirm that the predicted fixed points you found in part (a) are actually fixed points by computing their images under the function f. (c) (4 marks) Compute three iterates of the pointェ= 1/12, that is, compute f(1/12), fUU1/12)) and f(f(f(1/12))). What do you notice? (d) (4 marks) Start at the point = 1/12 and draw the iterates f(1/12), ful/12)) and f(f(f(1/12))) as a cobweb plot into the graph. Explain what this plot looks like and how it relates to what you noticed in part (c). You may use the attached extra copy of the graph for your answer if you wish

Answer 1

Answer:)

(a.)

We see that the graph of the map intersects the graph of the line y = x at three points, namely at (x,y) = (0,0), (0.5,0.5) and at (1,1). These are the fixed points we require.

(b.)

For x = 0:

$f(0) = 3 \cdot 0 = 0 \Rightarrow f(f(0)) = f(0) = 0 \cdots$

Thus, this is a fixed point of this map.

And, for x = 0.5:

f(0.5) f(f(0.5)) = f(0.5) = 0.5 2-3 . 0.5 = 0.5

This is again a fixed point of this map.

Finally, for x = 1:

$f(1) = -2 + 3\cdot 1 = 1 \Rightarrow f(f(1)) = f(1) = 1 \cdots$

Thus, this is also a fixed point of this map. and so all our claims are correct.

(c.)

We have:

12 4

We see that the function now has a period of two, that is it repeats after two iterations.

(d.)

We have the following cobweb plot (the bold path in black):

0:8 0:6 0:4 0.2 0.8 0.6 0.4 0.2